\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=\frac{1}{2}.\frac{49}{100}\)

\(=\frac{49}{200}\)

Giúp mk mấy bài nhaEdowa Conan

\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}\)

\(=\frac{49}{100}\)

\(\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)

= \(\frac{5}{2}-\frac{5}{4}+\frac{5}{4}-\frac{5}{6}+...+\frac{5}{98}-\frac{5}{100}\)

= \(\frac{5}{2}-\frac{5}{100}\)

= \(\frac{49}{50}\)

\(Q=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)

    \(=5\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

    \(=\frac{5}{2}.2.\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

    \(=\frac{5}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)

    \(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

    \(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{5}{2}.\frac{49}{100}=\frac{49}{40}\)

\(\Rightarrow Q=\frac{49}{40}\)

\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...........+\frac{1}{98.100}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

cho mình nha!

K = 4/2 - 4/4 + 4/4 - 4/6 + ....... + 4/2008 - 4/2010

K = 4/2 - 4/2010

K = 4016/2010 = 1/1003/1005

\(\frac{1004}{1005}\)

\(K=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)

\(K=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2002}-\frac{1}{2010}\right)\)

\(K=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(K=2.\frac{502}{1005}\)

\(K=\frac{1004}{1005}\)

\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(=2.\frac{502}{1005}\)

\(=\frac{1004}{1005}\)

Có gì ko hiểu thì cứ hỏi mình nha :)

Ta có: \(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(=2.2\frac{2}{4}+2.2\frac{2}{4.6}+2.2\frac{2}{6.8}+...+2.2\frac{2}{2008.2010}\)

\(=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(=2.\frac{1}{2}-2.\frac{1}{2010}\)

\(=1-\frac{1}{1005}\)

\(=\frac{1004}{1005}\)

\(\text{Ta có:}\) \(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(\Rightarrow\frac{1}{2}A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2008.2010}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2008}-\frac{1}{2010}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{2010}\)

\(\Rightarrow\frac{1}{2}A=\frac{502}{1005}\)

\(\Rightarrow A=\frac{502}{1005}:\frac{1}{2}=\frac{1004}{1005}\)

\(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+............+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)

\(B=\frac{4^2-2^2}{\left(2.4\right)^2}+\frac{6^2-4^2}{\left(4.6\right)^2}+..........+\frac{98^2-96^2}{\left(96.98\right)^2}+\frac{100^2-98^2}{\left(98.100\right)^2}\)

\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-...............-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)

\(B=\frac{1}{2^2}-\frac{1}{100^2}\)

\(B=\frac{1}{4}-\frac{1}{10000}\)

\(B=\frac{2500}{10000}-\frac{1}{10000}\)

\(B=\frac{2499}{10000}\)

Vậy B = \(\frac{2499}{10000}\)

Ta có :

\(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)

\(=\frac{12}{4.16}+\frac{20}{16.36}+...+\frac{388}{9216.9604}+\frac{396}{9604.10000}\)

\(=\frac{1}{4}-\frac{1}{16}+\frac{1}{16}-\frac{1}{36}+...+\frac{1}{9604}-\frac{1}{10000}\)

\(=\frac{1}{4}-\frac{1}{10000}< \frac{1}{4}\)

\(\Leftrightarrow B< \frac{1}{4}\)

B=\(\frac{12}{4.16}\)+\(\frac{20}{16.36}\)+...+\(\frac{396}{9604.10000}\)

Ta có:\(\frac{12}{4.16}\)=\(\frac{1}{4}\)-\(\frac{1}{16}\)

         \(\frac{20}{16.36}\)=\(\frac{1}{16}\)-\(\frac{1}{36}\)

            ...

Khi đó:B=\(\frac{1}{4}\)-\(\frac{1}{16}\)+\(\frac{1}{16}\)-\(\frac{1}{36}\)+...+\(\frac{1}{9604}\)-\(\frac{1}{10000}\)=\(\frac{1}{4}\)-\(\frac{1}{10000}\)<\(\frac{1}{4}\)

Vậy: B<\(\frac{1}{4}\)